Problem: Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}-8x+4y &= 2 \\ -2x+3y &= -4\end{align*}$
Solution: Begin by moving the $y$ -term in the second equation to the right side of the equation. $-2x = -3y-4$ Divide both sides by $-2$ to isolate $x$ $x = {\dfrac{3}{2}y + 2}$ Substitute this expression for $x$ in the first equation. $-8({\dfrac{3}{2}y + 2}) + 4y = 2$ $-12y - 16 + 4y = 2$ Simplify by combining terms, then solve for $y$ $-8y - 16 = 2$ $-8y = 18$ $y = -\dfrac{9}{4}$ Substitute $-\dfrac{9}{4}$ for $y$ in the top equation. $-8x+4( -\dfrac{9}{4}) = 2$ $-8x-9 = 2$ $-8x = 11$ $x = -\dfrac{11}{8}$ The solution is $\enspace x = -\dfrac{11}{8}, \enspace y = -\dfrac{9}{4}$.